查找在其他行上的其他两个字符串之间的所有字符串实例

Question

So I feel like I should know how to do this but I can't quite get it.

I'm trying to find all instances (in all files) where a string that ends with _START exists between two strings (that are normally on other lines) @GROUP and @END_GROUP

So there might be some code like this

// @GROUP GroupName OtherStuff
#define GROUPNAME_START 1
#define GROUPNAME_FOO 2
.... (more defines)
#define GROUPNAME_END 10
// @END_GROUP

#define GROUPTWO_START 1
// @GROUP GroupTwo MoreStuff
#define GROUPTWO_FOO 2
.... (some defines)
#define GROUPTWO_BAR 70
// @END_GROUP

And I would want to match the first group (really just the line with _START, but everything would be ok) but not the second group or the _START line that is outside of the @GROUP comments.

I figure using grep for this would be the best way to search through all the files, but I can't quite get the regex needed. Thanks for the help.

edit: My bad for not making it clear that I want to be able to search through files in multiple directories at the same time, doing the same as a grep -r "foo" * . Answers have been good, I just didn't make that clear.

edit2: Multiple great answers each solved it in a slightly different way and I really don't know which one would be best. I marked the one who responded first, but anyone looking at this should be sure to check out all the answers, one might be better for your problem.

Answer 1

grep only sees one line, so it doesn't know whether it's between the group comments or not. sed can use addresses, though:

sed '/@GROUP/,/@END_GROUP/!d' input_file | grep '_START'

! negates the addresses, d deletes a line, ie we're telling sed to remove lines that are not between the group comments. grep then operates only on the "interesting" lines.

To make it work for subdirectories, too, add find to the toolbox:

find /path/to/dir -type f -exec sed '/@GROUP/,/@END_GROUP/!d' {} + | grep '_START'

Or, if the group comment could appear without the corresponding END, use a slower but safer

find /path/to/dir -type f -exec sed '/@GROUP/,/@END_GROUP/!d' {} \; | grep '_START'

Or, let xargs operate on the output of grep -l :

grep -lr @GROUP /path/to/dir | xargs sed '/@GROUP/,/@END_GROUP/!d' | grep '_START'

Note: If your filenames contain spaces, it wouldn't work.

Answer 2

With awk you can use null RS and do all that in single search:

awk -v RS= '/@GROUP.*_START.*@END_GROUP/' file
// @GROUP GroupName OtherStuff
#define GROUPNAME_START 1
#define GROUPNAME_FOO 2
.... (more defines)
#define GROUPNAME_END 10
// @END_GROUP

Answer 3

This is a job for sed , using its address syntax:

#!/bin/sed -f

/@GROUP/h  # store the @GROUP line

/@GROUP/,/@END_GROUP/{
/_START/{
g  # retrieve the @GROUP line
n  # print it and continue
}
}

# otherwise, delete the line and continue
d

It's a little bit complicated by the nested blocks, but what this does is: within @GROUP .. @END_GROUP , then for any line matching _START it will print the previously found @GROUP line thus (using your example):

$ ./group.sed group.data 
// @GROUP GroupName OtherStuff

Is that what you're trying to achieve?

Edit : It's not what you asked for - you just want the _START line, not the @GROUP line. Well that's much easier:

#!/bin/sed -nf
/@GROUP/,/@END_GROUP/{
/_START/p
}

Addendum : Since you now ask for recursive directory searching, you can use find as described in other answers:

find . -type f -print0 | xargs -0 ./group.sed --separate

(I've used the GNU sed --separate argument here to protect against any file having the group start but missing the group end line).